Download Technote 7 Using Op Amps Successfully

Technote 7
August 2006
Revised 05/08/08
Using Op Amps Successfully
Tim J. Sobering
SDE Consulting
[email protected]
© 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 2
Using Op Amps Successfully
Op Amps are extremely useful and fairly simple devices. However, the basic ideal model
presented in typical Circuit Theory and Electronics classes can trick most students into
believing they understand these devices. In reality, while you have been taught more than
you realize, what is needed is for you to make a few more connections between the facts and
methods you have learned.
Let’s start by reviewing the basics. Op Amps are high-gain DC coupled amplifiers usually
intended to amplify signals over a wide frequency range. They have a differential input
and a single-ended output. This is illustrated in Figure 1.
V1
+
Va
V2 -
Av
Vo = AvVa = Av(V1 - V2)
Figure 1. Op Amp in an “open-loop” configuration
You have already been exposed to a set of ideal assumptions that make it quite easy to
analyze an Op Amp circuit. For example, you have been taught that Av, the open-loop
gain1, is infinite. Initially this sounds confusing because if the gain is infinite, shouldn’t the
output voltage be infinite as well? How is this device useful? In reality, Op Amps work
because they are always configured so the output will “drive” the input differential voltage
nearly to zero. This is called negative feedback and is the key to making Op Amps so
useful. In your classes you were probably taught about “virtual ground2” or told that the
voltage between the Op Amp’s inputs is zero. This is what was being described to you.
Keep this concept in mind as it helps in the analysis, as I will show later.
Figure 2 illustrates a more detailed but still simplified model for a voltage feedback3 Op
Amp. Table 1 summarizes the remaining ideal Op Amp assumptions. Some of these terms
may be unfamiliar to you so I will try to clarify each of these characteristics. What is
Av ≈
Va
Ri
Vo
Va
AvVa
(1)
Ro
Vo
Ideal Characteristic
No current flows in the inputs
No output voltage for zero input
Voltage gain is infinite
Output impedance is zero
Frequency response is flat and
infinite
Slew Rate is infinite
Infinite CMRR and CMR
PSRR is infinite
Figure 2. Op Amp Equivalent
Circuit
Noise free
Circuit Effect
Ibias(+,−) = 0 or Zin = ∞
Vos = 0
Av = ∞, Va = 0
Zout = 0
BW = ∞ or Delay = 0 and
GBW = ∞
BW = ∞ or Delay = 0 and
GBW = ∞
Unlimited input voltage range.
Differential response only
Immune to PS offset and
variations
Output due to signal only
Table 1. Ideal Op Amp Characteristics
This is the gain when there is no feedback to “close the loop”. Depending on the reference this can also be designated as AOL
or possibly something different. Use caution to avoid confusion as it is a critical parameter.
2 I do not like this term as “ground” really only applies in the inverting Op Amp configuration. It is better to think in terms of
the inputs being driven to the same potential.
3 There are two types of Op Amps – Voltage Feedback and Current Feedback. They are similar but differ in critical points.
Current Feedback Op Amps are not as common and generally used in high speed applications. They will not be covered in
this document.
1
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 3
important at this point is that the ideal Op Amp assumptions make analyzing an Op Amp
circuit trivial. Keep in mind that you should always start an analysis using the ideal
assumptions. If the circuit does not work using an ideal analysis, there is no point in
performing a more detailed analysis. It should be obvious that the ideal characteristics
listed will be replaced later with real characteristics. However, often each characteristic
can be examined individually to determine the effect on the actual circuit.
Figures 3 and 4 show the inverting and non-inverting Op Amp configurations. The
impedances Zi and Zf “close the loop” and determine the “gain” of the circuit (in addition to
other characteristics). Looking at Figure 3, the output voltage is “fed back” to the inverting
input terminal. This provides the negative feedback discussed above4. Think in these
terms: as the output voltage increases, the voltage at the inverting input (V2 in Figure 1)
increases (remember Superposition?), which causes the difference V1-V2 to decrease, which
causes the output to decrease. This is negative feedback.
Zf
Vi
Zf
Zi
Zi
Va
Av
Vo
Va
Av
Vo
Vi
Figure 3. Inverting Op Amp Configuration
Figure 4. Non-Inverting Op Amp Configuration
How does this help? Again referring to Figure 3, because there is negative feedback, the
output can drive the input differential voltage Va to zero. Then the signal, or closed-loop,
gain can be determined by applying KCL at the inverting input terminal.
Vi + Va Vo + Va
+
=0
Zi
Zf
(2)
Keep in mind that we are assuming that the Op Amp has infinite input impedance, which
means no current flows into either input. Also, we are assuming that there is no DC error
at the output (offset voltage VOS = 0 V). And, if the open-loop gain is infinite, Va = 0 V.
Equation (2) then simplifies to:
Zf
Vo
=−
Vi
Zi
(3)
The ideal analysis of the circuit in Figure 4 is similar. However, if Va = 0 V, the voltage at
the inverting input node must be Vi and the analysis yields:
Zf
Vo
= 1+
Vi
Zi
(4)
4 The feedback connection in an Op Amp does not have to be made to the inverting input terminal. What is important is that
the feedback voltage drives the output in the opposite direction. If there is “something” in the feedback look that inverts the
phase, the feedback would need to be to the non-inverting input.
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 4
What this means is that the (ideal) signal gain of an Op Amp circuit is determined only by
the feedback network. For now I will skip over discussing the effects of non-infinite open
loop gain (it will be discussed in detail below) and move on to the other ideal assumptions.
However, notice that while the gain of the inverting configuration can assume (nearly) any
value, the non-inverting configuration is limited to gains greater than one. In fact, a
special case for the non-inverting configuration is when Zi → ∞. This is called a unity gain
buffer or non-inverting voltage follower.
The input offset voltage (VOS) is a DC error in the Op Amp and equals the voltage that
needs to be added to the input to result in a zero volt output. It is usually a few microvolts
to a few millivolts. Because of the high open loop gain of an Op Amp, this is never
measured directly but instead, the Op Amp is configured as a non-inverting voltage follower
(Figure 4 with Zi = ∞ and Zf = 0 Ω) with Vi grounded. In this configuration, the output
voltage equals VOS. If you want to determine the effect of VOS on the output voltage, it can
be modeled as a voltage source in series with either the inverting or non-inverting input pin
of the device. Eliminating this parameter by assuming it is zero slightly simplifies an
analysis. However, the easiest way to include offset voltage in the analysis is to use
superposition and analyze it separately. Thus, the contribution to the output voltage due to
VOS for the amplifiers in either Figure 3 or Figure 45 is given by:
⎛ Rf
Vo V = VOS ⎜⎜1 +
OS
Ri
⎝
⎞
⎟⎟
⎠
(5)
Resistances are used as this is strictly a DC error. The polarity of VOS is usually ignored as
it is a function of the design and manufacturing and typically cannot be predicted.
The input bias current is the current flowing into or out of the input terminal. Bias
currents are modeled as current sources from each input pin to ground and can range from
femtoamps (for electrometer type Op Amps) to microamps. The direction of current flow is
also a function of the design and manufacturing and cannot be predicted without some
additional knowledge of the device. Again, eliminating bias currents makes the ideal
analysis simpler, but it can be easily included again by using superposition. Referring to
Figures 3 and 4 and assuming the bias current flows from ground to the input pin, the
output voltage contributions are given by:
Vo
I BIAS −
Vo
= − R f I BIAS −
(6)
=0
(7)
I BIAS +
Again, resistances are used because this is a DC error. The contribution due to the noninverting input bias current is zero because there is no resistance present for it to develop a
voltage across. This is certainly the case in Figure 3. In Figure 4 this is problematic as Vi
is most likely generated by a source having a source impedance. In this case, the
contribution becomes:
Vo
I BIAS +
⎛ Rf
= I BIAS + Rs ⎜⎜1 +
Ri
⎝
⎞
⎟⎟
⎠
(8)
5 When you apply superposition to compute the output contribution due to VOS to the amplifiers in Figures 3 or 4, the circuits
become identical.
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 5
This is the voltage induced on the source resistance times the DC gain of the amplifier.
Figure 5 shows the Op Amp model including sources representing offset voltage and input
bias currents. While it is starting to look more complicated, keep in mind that
superposition can be used to greatly simplify calculating the contributions from each of
these sources.
Vos
+Ibias+
Va
Ri
AvVa
Ro
Vo
IbiasFigure 5. Op Amp model including common parameters.
It is possible to compensate for either input offset voltage or bias currents errors. However,
caution must be used or additional problems can be created. In the case of VOS, some Op
Amps have offset null terminals and the data sheet will recommend a resistor network,
usually including a potentiometer. The intention is to sum a voltage into the input stage
and eliminate the offset voltage. My best advice is to never use these pins. First,
potentiometers, while convenient, are notoriously bad devices having poor temperature
characteristics and poor stability. Usually, these networks severely degrade the drift of the
Op Amps (errors vs. temperature). As an alternative, it is possible to build an external
network to sum a voltage into the circuit and eliminate the offset. However, the source
must be very stable and temperature compensated. Usually this is not worth the effort. As
a general rule, if VOS is a problem, pick a better Op Amp.
Similarly, it is possible to compensate for input bias current. In Figure 3, if a resistor is
added between the non-inverting input and ground with a value equal to Rf ⎜⎜Ri, the output
voltage due to the bias currents is eliminated. This can be shown using Equation (8) and
letting Rs = Rf ⎜⎜Ri.
Vo
I BIAS +
⎛ R f Ri
= I BIAS + ⎜
⎜R +R
i
⎝ f
⎞⎛ R f
⎟⎜1 +
⎟⎜
Ri
⎠⎝
⎞
⎟⎟ = I BIAS + R f
⎠
(9)
This is equal to but of opposite polarity to the contribution due to IBIAS- so they cancel.
However, the addition of a bias current cancellation resistor can increase the noise at the
output of the Op Amp. The problem is fixed by adding a capacitor in parallel with the bias
current cancellation resistor to limit the noise bandwidth6.
The problem is more difficult with the circuit in Figure 4. Because it is likely that the
source impedance driving Vi is low, it is difficult to make Rs = Rf ⎜⎜Ri. It is possible to add a
You are not expected to understand this at this time. “Noise” will have to be addressed later. Just keep in mind that you
should not add a bias current compensation resistor without a parallel capacitor (1 µF is usually sufficient).
6
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 6
series resistor at the non-inverting input such that Rs + Rseries = Rf ⎜⎜Ri, but there is no good
way to add capacitance to prevent the noise performance from being degraded.
Finally, there is one more consideration. The cancellation techniques discussed above
assume the bias currents are both flowing in the same direction. However, it has become
more common for manufacturers to add input circuitry to their Op Amps to try and reduce
the magnitude of the bias currents. This can cause the bias currents to flow in different
directions. The specification to watch is the input bias current offset, which is defined as
the difference between the inverting and non-inverting bias currents. If the bias current
offset is small compared to the bias current, it is safe to assume the bias currents flow in
the same direction. However, if the magnitudes of the bias current and bias current offset
are similar, there is no way to be sure that the inverting and non-inverting bias currents
have the same direction and the techniques discussed above can actually make the problem
worse. Generally, the best solution is to pick a better Op Amp with a lower bias current.
The remaining ideal assumptions from Table 1 are a little less obvious but useful none the
less. For example, zero output impedance means that the Op Amp operates independently
of the load impedance. In reality, the Op Amp has a finite drive capability and also the load
impedance can be critical to the stability of the amplifier. This will be discussed later when
open loop gain is revisited. The frequency response clearly cannot be flat and infinite
(although with care the response of the amplifier can be flat out to a cutoff frequency). Op
Amps have finite small signal bandwidth and in fact, this is where things get very
interesting (again, more on this later). Slew rate is a large signal parameter and a measure
of how fast (dVo/dt) the output voltage can change. Op Amps vary widely in this respect
from 0.5 V/µs to over 1000 V/µs.
CMR is common mode input range. CMRR is common mode rejection ratio. Common mode
(CM) voltage is frequently misunderstood. Referring to Figure 1, the common mode voltage
is defined as
VCM =
V1 + V2
2
(10)
As noted above, Op Amps are differential amplifiers; they are intended to respond only to
the voltage difference between the input terminals. In Figures 3 and 4 above, this is how
we analyzed the circuits. Note however that there is an important difference in these
circuits. In Figure 3, the common mode voltage is zero. In Figure 4, VCM = Vi. In the ideal
model we assume this will have no effect. In reality, there are two considerations. First,
there is a limit to how close the input can get to the power supply rails. This is the CMR
and occurs because the input transistors of the Op Amp must remain properly biased for
linear operation. Second, there is a contribution to the output voltage due to the common
mode input voltage. This is the common mode gain; more commonly the reciprocal is used
and referred to as the CMRR. CMRR is a function of frequency and is a measure of how
well the Op Amp rejects common mode signals. Generally, CMRR is high (low CM gain) at
low frequencies but drops –20 dB/decade as frequency increases. Thus, high frequency CM
signals can be a problem. In this respect, the amplifier in Figure 3 will perform “better”
than the amplifier in Figure 4.
This is an opportune point to digress into a discussion of “ground”. Clearly “voltage” must
be measured between two points. “Ground” is the word used for the “reference” potential in
a circuit and is generally assumed to be zero volts. Note however that Op Amps do not
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 7
have a ground terminal. Ground is implied. This is easy to ignore. After all, most
amplifiers, at least the older ones, operate from ±15 V or ±12 V rails and are biased
symmetrically above and below ground. “Ground” is than half way between the rails and
everything works as long as your signals remain a volt or two inside the rails. However,
what happens when you use a single supply Op Amp? If the amplifier in Figure 3 is
operated from 5V to ground, caution is required because the non-inverting terminal is
connected to ground. First, this will likely violate the common mode input range for the
device because there is no “headroom” to bias the input circuitry properly7. Second, this is
an inverting amplifier. However, as the output cannot go below ground, it will only work
when Vi is negative. Finally, the amplifier will not work for small input voltages because
the output cannot get close enough to ground to drive the inverting input to zero volts.
Operating a single supply Op Amp in this manner is a very common error because there is
a generally poor understanding of common mode input range.
There are a number of solutions to this problem and a great deal of literature on how to use
single supply Op Amps. I will not present those solutions in detail here but generally they
involve redefining “ground” by applying a reference voltage to the non-inverting terminal or
by summing a fixed reference into the inverting input node. Caution must be used to
prevent this reference from being amplified. One approach is to create a reference voltage
at the midpoint of the supplies and redefine the circuit “grounds” to that voltage. This
works well. The points are that “ground” is a concept that the designer has to address,
there can be multiple “grounds” in a system, and keep in mind that Op Amps have a limited
operating range that is bounded by (and likely inside) the supply rails.
Back to the ideal assumptions in Table 1; PSRR is power-supply rejection ratio and is
similar to CMRR. The assumption is that variations on the power supply rails do not affect
the output voltage, i.e. infinite PSRR. In fact, this is why Op Amps are frequently drawn
without power pins. We generally assume that as long as the Op Amp has sufficient supply
voltage to properly bias the internal circuitry, the power pins aren’t important. This is very
dangerous as properly powering an Op Amp is critical to the performance of the amplifier.
This is why power supply decoupling8 is necessary. In reality, the power pins are just
another input and have to be engineered accordingly.
Finally, the general assumption is that the Op Amp is a noise free device. Noise is another
topic that will not be addressed in detail at this point but I will touch on the basics9. Op
Amps have three noise sources associated with them, a voltage noise source and a current
noise source on each input. In fact, the noise model looks just like Figure 5 with VOS
replaced with en (the Op Amp voltage noise) and Ibias+ and Ibias- replaced with in- and in+ (the
Op Amp current noise sources). The difference is that these are now AC sources. Again,
the output contributions can be determined using superposition; however the analysis is
more complex as these noise sources have a spectral density that is a function of frequency.
7 There are rail-to-rail Op Amps available, but read the datasheet carefully as rail-to-rail typically means operation within 50
mV (or more) of the supply rail. Also, a rail-to-rail output Op Amp is not necessarily a rail-to-rail input Op Amp, and visa
versa. However, some rail-to-rail Op Amps can be operated in this manner provided Vi stays below zero volts.
8 Decoupling is a topic onto itself and will not be discussed in detail here. It involves using capacitors and possibly resistors or
inductors to “decouple” the component from the power bus and prevent components from “talking” to each other over the
power bus. Many datasheets do not show decoupling networks or only show minimal decoupling. That is because when a
component is used by itself, little or no decoupling is needed. However, the manufacturers assume the designer will
“engineer” the power distribution and grounding for a system and decoupling networks are implied.
9 A good starting point is “Noise Analysis in Operational Amplifier Circuits”, Texas Instruments, SLVA043A, 1999,
“Operational Amplifier Noise Prediction”, Intersil, AN519.1, November 1996, and “Noise Analysis Of FET Transimpedance
Amplifiers”, Texas Instruments, AB-076, 1994.
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 8
In addition, to get a true indication of the noise performance of an amplifier, the noise
contributions from the resistors must be included. Resistors have a noise voltage given by
VnR = 4kTRΔf
(11)
Where R is the value of the resistor, T is the temperature, k is Boltzmann’s constant, and Δf
is the noise bandwidth10. To properly analyze the noise performance of an Op Amp, it is
necessary to know the details of the frequency response.
Now let’s take a closer look at one of the key assumptions made above, infinite open loop
gain. This assumption enabled us to quickly and easily analyze the circuits in Figures 3
and 4. Of course, the reality is that open loop gain is finite and is a function of frequency.
Op Amp open loop gain is often approximated using a single pole at ~10Hz. Gains at DC
can be 80 dB to 140 dB, depending on the speed or precision of the device, and a single pole
causes a 90° phase shift. I’m jumping ahead a little, but problems don’t occur until the
phase shift11 starts approaching 180°. Op Amps actually have several poles in their
frequency response because internally they can have 2 to 5 amplifier stages. How much of
an effect this has depends on the location of the other poles. A fully compensated Op Amp,
also called unity-gain stable, is one in which the 2nd pole has been shifted up in frequency to
the point where the phase shift is << 180° when Av = 0 dB12. This point is known as the
unity-gain bandwidth or gain-bandwidth product (GBW or fT).
Let’s introduce some more definitions. The feedback factor (β) is defined as the fraction of
the Op Amp output that is fed back to the input. This is a critical parameter and it will be
shown below that much about the Op Amp’s AC performance and stability can be predicted
from this value. Using the amplifier from either Figure 3 or Figure 4 and removing the Op
Amp for simplicity, the feedback factor can be defined as (reference Figure 6):
β=
Zi
V2
=
V1 Z i + Z f
(12)
V1
Zf
V2
Zi
Figure 6: Op Amp Feedback Network
Next, we define the term “noise gain” as the reciprocal of β.
Noise _ Gain =
1
β
=
Zi + Z f
Zi
= 1+
Zf
Zi
(13)
10 Equivalent Noise Bandwidth corresponds to a rectangular, or “brickwall” filter of bandwidth Δf with has the same noise
power as the circuit bandwidth, i.e. the integrals over frequency are equal. For a single pole response typical in an Op Amp
circuit, this is given by Δf = f3dB(π/2).
11 Actually it is the phase shift associated with the loop gain Avl that is important. Avl = Av only when Zf = 0 and Zi = ∞. This
will be discussed in more detail later.
12 There may be a more formal definition out there but manufacturers vary a lot. As you will learn later, there are numerous
effects that contribute to the phase and ultimately to stability of the amplifier. Many manufactures aim for a <45° phase
margin for an Op Amp (difference between the phase at Av = 0 dB and 180°). While this is a useful design consideration, it is
not the complete story.
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 9
Finally, we define the loop gain, Avl, as the gain around the feedback loop, as illustrated in
Figure 7. The loop gain is given by
Avl =
V0
Zi
=
Av = Av β
V1 Z i + Z f
(14)
V1
Zf
Zi
Va
Av
Vo
Figure 7: Op Amp Loop Gain
First, note that the noise gain is the same as the closed loop signal gain of a non-inverting
amplifier configuration. Second, note that β and Avl are the same regardless of which
amplifier configuration, inverting or non-inverting, is examined. This is a good point to
stop and reread this material because this is a critical observation. This means the
performance of an amplifier circuit (stability, bandwidth, noise, overshoot, ringing, etc.)
depend only on the open loop gain, Av, and the feedback factor, β, and are otherwise
independent of the amplifier configuration.
Returning to our discussion of open loop gain and bandwidth, the small signal bandwidth,
or frequency response, or closed loop bandwidth13 of an amplifier is the frequency where the
signal gain is 3 dB below the passband gain. It is a property of voltage feedback14 Op Amps
that the closed loop bandwidth is given by fT divided by the circuit noise gain, or fCL = fTβ.
Graphically, the above is summarized in the magnitude plot in Figure 8. Note that when
plotted in dB, the difference between the open loop gain (Av) and the noise gain is the loop
gain (Avβ). This will be very useful later.
Gain (dB)
OPEN LOOP GAIN
-20 dB/DECADE
Avβ
fCL = CLOSED-LOOP BANDWIDTH
NOISE GAIN (1/β)
fT
0 dB
fCL = fTβ
LOG f
Figure 8: Op Amp Open Loop Gain and Noise Gain
Sorry to use so many terms but we engineers curse ourselves by frequently using numerous terms to describe the same
characteristic or one term to describe multiple characteristics. Consider yourself warned.
14 Current feedback amplifiers do not follow the GBW relationship.
13
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 10
Now let’s look closely at what happens if Av is not infinite. Referring to the inverting
amplifier in Figure 2 above, applying KCL at the inverting node, and noting that Vo = AvVa,
we can derive an expression for the closed loop inverting gain Avcl(inv).
Vi + Va Vo + Va
+
=0
Zi
Zf
(15)
Vi
Vo
V
Vo
+
+ o +
=0
Z i Av Z i Z f Av Z f
(16)
⎡ 1
1
1
Vo ⎢
+
+
⎢⎣ Av Z i Z f Av Z f
Vo
1
= Avcl (inv ) = −
Vi
Zi
Avcl (inv ) = −
⎤
Vi
⎥=−
Zi
⎥⎦
1
1
1
1
+
+
Av Z i Z f Av Z f
Av Z f
(17)
(18)
(19)
Z i + Z f + Av Z i
Which can be rewritten as:
⎛
⎜
Zf ⎜
Av
Avcl (inv ) = −
⎜
Zi Zi + Z f
⎜⎜
+ Av
⎝ Zi
Recognizing that β =
⎞
⎟
⎟
⎟
⎟⎟
⎠
(20)
Zf
Zi
and defining α =
, this equation can be simplified to:
Zi + Z f
Zi + Z f
⎞
⎛
Zf ⎜
⎟
1
α
Avcl (inv ) = −
=−
⎟
⎜
Zi ⎜ 1+ 1
β
Av β ⎟⎠
⎝
⎞
⎛
⎟
⎜
1
⎟
⎜
⎜1+ 1 A β ⎟
v
⎠
⎝
(21)
Note that as Av → ∞
Avcl (inv ) = −
Zf
Zi
=−
α
β
(22)
Similarly, for the non-inverting amplifier in Figure 3 we obtain an expression for the closed
loop non-inverting gain Avcl(non).
⎛ Zf
Avcl (non ) = ⎜⎜1 +
Zi
⎝
⎞
⎛
⎞
⎛
⎟
⎟ 1⎜
⎞⎜
1
1
⎟⎟⎜
= ⎜
⎟
⎟
⎠⎜ 1 + 1 A β ⎟ β ⎜ 1 + 1 A β ⎟
v
v
⎠
⎝
⎠
⎝
Copyright © 2006 Tim J. Sobering. All rights reserved.
(23)
Using Op Amps Successfully
Page 11
Again as Av → ∞
Avcl (non ) = 1 +
Zf
Zi
=
1
(24)
β
Notice that in both cases, the close loop gain is expressed as the ideal gain times an error
term that is strictly a function of the loop gain Avβ. The gain error can be computed from:
Gain _ Error = 1 −
1
1+ 1
(25)
Av β
Note that:
1
1+ 1
=
Av β
Av β
1
≈ 1−
1 + Av β
Av β
(26)
Thus, the gain error can be simplified and expressed as a percentage using:
ε (%) ≈
100
Av β
(27)
Figure 8 shows that Avβ (shown graphically as the difference between the open loop gain
and the noise gain is the loop gain) decreases with frequency. This is why high open loop
gain is important and also why it is often necessary to select an Op Amp with high open
loop gain and/or a higher then expected bandwidth. Otherwise the gain error given by
Equation (27) can be large. Stability can also be impacted.
This brings us to the main topics of this discussion: feedback factor (β) and loop gain (Avl).
These terms are critical in assessing the AC performance and stability of any Op Amp
configuration. Figures 9 and 10 show inverting and non-inverting amplifiers with signal
gains of -2 and +2, respectively. For the inverting amplifier in Figure 9, β = 1/3 and for the
non-inverting amplifier in Figure 10, β = 1/2. This corresponds to noise gains of 3 and 2,
respectively.
2k
Vi
1k
1k
1k
Av
Vo
Av
Vo
Vi
Figure 9. Inverting Gain of -2
Figure 10. Non-Inverting Gain of +2
This means that while these amplifiers have the same signal gain15, the non-inverting
amplifier has a higher bandwidth because it has a lower noise gain! Remember, bandwidth
is given by fTβ.
15
Yes, I’m ignoring the minus sign. In a great many cases a 180° phase shift can be ignored.
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 12
Another brief aside, using the terminology presented above, we can represent the inverting
and non-inverting as control system blocks as shown in Figures 11 and 12. This notation is
commonly used in a number of texts.
Vi
-α
Σ
Va
Av
Vo
Vi
Σ
Av
Vo
β
β
Figure 11. Inverting Amplifier Block Diagram
Va
Figure 12. Non-Inverting Amplifier Block Diagram
Now we can finally start discussing stability. The Nyquist Criterion states that a circuit
will oscillate if the loop gain Avβ = 1∠180°. From the discussion above, we already know
that Av contributes at least a 90° phase shift and in many cases we may not have much
leeway to select a different amplifier. So the stability of the amplifier will depend largely
on the characteristics of β. However, it is not sufficient just to insure that the phase of the
loop gain is less than 180°. Yes, the amplifier might not spontaneously oscillate, but as Avβ
approaches 180°, the amplifier will exhibit increasing overshoot and ringing in its step
response. Generally, 135°, or a phase margin of 45°, is a good starting point. However,
more phase margin may be required, or a lower phase margin may have to be tolerated,
depending on the individual application.
But wait a minute. If you are using an Op Amp with purely resistive feedback, β doesn’t
contribute a phase shift. So provided the open-loop gain doesn’t introduce an excessive
phase shift, everything works, right? Wrong. As with most things, it is what you don’t see
that will ruin your day.
There are a number of “components” missing from the circuits in Figures 3 and 4. Above
we alluded to a “source resistance”, a characteristic of whatever is driving the input of the
amplifier. In reality this is an impedance which can change the loop gain. The amplifier
also drives a load, so load impedance has to be considered. A real Op Amp has a finite
output impedance which interacts with the load to alter the loop gain. An Op Amp also has
differential and common mode input impedances (typically modeled as high value resistors
in parallel with a small capacitance). These also affect the loop gain. Finally, every circuit
has stray capacitances that are a function of the circuit construction methodology. These
can be minimized but not eliminated.
At this point the simple Op Amp circuits we discussed above are getting pretty ugly. As a
general rule, if you have a high source impedance, a large capacitive load, large feedback
resistors, or want a high frequency response, you need to take a close look at the loop gain16.
You can analyze the circuit by hand or use one of the many graphical shortcuts to
determine the loop gain, but this is time consuming (and can be quite intimidating even
though you really only need the loop gain at a single frequency). You can build the circuit
and try and measure the loop gain, but I am not a big fan of iterative design. The
alternative I use is to simulate the amplifier using PSpice or some other circuit simulation
package.
Circuit simulation software is dangerous. There is a significant tendency to blindly rely on
what the computer predicts without having a good understanding of the circuit or the models
This comes down to the question “how big is big?” Unfortunately too many things are interrelated to give set guidelines.
You will have to develop your own experience.
16
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 13
being used. A cute example is the uA771. Odds are that the macromodel in your software
package includes a value of Rp = 10.00E-3. Rp is the power dissipation resistance and is
only used to simulate the current drawn from the power supply. Assuming supply rails of
+/- 15V, this model will draw 90 kW. Obviously this is a typographic error and it should
probably be 10.00E3. The point is that macromodels can contain errors and, more
importantly, only model a limited set of parameters for a given amplifier. Sometimes they
are quite basic and leave out PSRR, CMRR, input and output impedances, etc. Be very
skeptical of any model and study the macromodel to learn its limitations.
Obtaining the open loop gain from a simulation is problematic. Simply connecting the Op
Amp in an open loop configuration will likely yield erroneous results, especially with newer
macromodels. This is because as manufacturers have improved their macromodels, for
example by adding additional structures to the inputs to more accurately simulate bias
current compensation, it is important that the inputs be properly biased in order to
correctly simulate the AC response of the Op Amp. With that said, there is a very good
application note17 from Harris Semiconductor that gives a reliable method for obtaining
open loop gain, loop gain, and feedback factor for a given amplifier configuration. This is
shown in Figure 13. R1 and R2 provide a DC bias for the Op Amp input. However, C1
effectively shorts the inverting input to ground for AC signals. Thus, at DC the Op Amp is
properly biased and for AC R2 appears as a negligible load on the output. So the probes on
U1 pin 6 give the magnitude (in dB) and phase of Av directly (because our input signal is
1V).
Now that we have Av, we need to obtain the feedback factor, β. In Figure 13 below, Rf is the
feedback resistor and Ri the input resistor. The divider formed by Rf and Ri represents the
feedback factor β. The point where these two resistors connect yields Avβ. Because the
markers are in dB, the difference between the markers is the noise gain, 1/β (in dB). With
1k and 10k resistor values, the noise gain is 11.
I added components18 to the schematic to model input capacitance, Cin, load capacitance,
Cload, and feedback capacitance, Cf. You can easily add other components to introduce stray
impedances or source impedances if required. Using this approach, you can fairly
accurately determine the phase margin for your design. You should still study the Op Amp
macro model to fully understand the limits of your simulation.
2
-
4
V3
1Vac
0Vdc
-V
V1
+15Vdc
V2
-15Vdc
0
OS2
OUT
V-
+V
+
OS1
U1
uA741
Cf
0.01pF
Av
6
1
Av B
Rf
1MEG
VP
Rload
10k
Cload
0.01pF
VP
Rg
1MEG
Cin
0.01pF
-V
0
0
VDB
5
VDB
3
V+
7
+V
0
0
0
C1
1000GF
0
0
R1
20k
R2
20k
Figure 13. Open Loop Gain Simulation
17
18
“PSPICE Performs Open Loop Stability Analysis (HA5112)”, Harris Semiconductor, November, 1996, AN9536.1.
I leave these components in my simulation even if they are not required as a value of 0.01pF is so small as to not matter.
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 14
Using this model, we can now take a quick look at some apparently minor but in reality
significant issues. First, let’s simulate the circuit using Rf = Ri = 1 kΩ, Ci = 20 pF, and leave
Cload = Cf = 0.01 pF. Then let’s change Rf = Ri = 1 MΩ leaving the other values unchanged.
The results are shown in Figures 14 and 15 below.
120
100
50
-0
-50
-100
-150
-180
1.0Hz
VDB(AvB)
VP(AvB)
10Hz
VP(Av)
VDB(Av)
100Hz
VDB(Av)-VDB(AvB)
1.0KHz
10KHz
100KHz
1.0MHz
10MHz
100MHz
1.0MHz
10MHz
100MHz
Frequency
Figure 14. μA741 frequency response, gain of 2, 1 kΩ
120
100
50
-0
-50
-100
-150
-180
1.0Hz
VDB(AvB)
VP(AvB)
10Hz
VP(Av)
VDB(Av)
100Hz
VDB(Av)-VDB(AvB)
1.0KHz
10KHz
100KHz
Frequency
Figure 15. μA741 frequency response, gain of 2, 1 MΩ
The open loop gain (yellow) and phase (blue) are the same, as expected. However the loop
gain (green) and phase (red), and noise gain (magenta) change significantly. With the 1 kΩ
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 15
resistors, the phase margin is ~70°. With the 1 MΩ resistors, the phase margin has
dropped to less than 10°. As it turns out, 20 pF is a very reasonable value for the input
capacitance (differential plus common mode) of an Op Amp. If you construct your circuit on
a white prototyping board, the stray capacitance introduced can be much higher.
Figures 16 and 17 show the effect on the step response of the amplifiers. With a phase
margin near 70°, the step response is well behaved. However, as the phase margin nears
180°, ringing dominates the response (although the oscillation is not sustained).
2.4V
2.0V
1.6V
1.2V
0.8V
0.4V
0V
0s
V(V4:+)
10us
V(U1:OUT)
20us
30us
40us
50us
60us
70us
80us
90us
100us
80us
90us
100us
Time
Figure 16. μA741 transient response, gain of 2, 1 kΩ
6.0V
5.0V
4.0V
3.0V
2.0V
1.0V
0V
0s
V(V4:+)
10us
V(U1:OUT)
20us
30us
40us
50us
60us
70us
Time
Figure 17. μA741 transient response, gain of 2, 1 MΩ
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 16
Refer back to Figures 14 and 15 and note the amplitude responses. For the 1 kΩ resistors,
the noise gain is flat out to the intersection with the open loop gain. With the 1 MΩ
resistors, there is a significant peak in the noise gain19 (~20 dB for this case). Recall that
the closed loop bandwidth is fCL = fTβ. What is the closed loop bandwidth when there is a
peak in the noise gain? By this strict definition, the bandwidth of the 1 MΩ amplifier is
~100 kHz, 5 times less than the amplifier built using the 1 kΩ resistors. This is true even
though both amplifiers have the same noise gain at 500 kHz. This makes sense if you think
in terms of “when does the amplifier stop being an amplifier?” The answer is that the loop
gain is equal to or less than 0 dB. Think of it this way: when the loop gain is < 0 dB the
amplifier can no longer drive the input differential voltage to zero (a key assumption from
above). Therefore, at this point all semblance of ideal behavior is gone.
There is one additional effect that is worth illustrating before moving to a discussion of
poles and zeros. Figure 18 repeats the AC simulation using Rf = Ri = 1 MΩ, Ci = 20 pF, but
this time includes Cf = 5 pF. Compare this to Figure 15 and notice that we now have a
phase margin of nearly 80°, ~2.5 times the bandwidth, and > 6 db reduction in peaking.
This illustrates why Bob Pease20 states that you always include a small feedback capacitor
“unless you can prove you don’t need it”.
120
100
50
-0
-50
-100
-150
-180
1.0Hz
VDB(AvB)
VP(AvB)
10Hz
VP(Av)
VDB(Av)
100Hz
VDB(Av)-VDB(AvB)
1.0KHz
10KHz
100KHz
1.0MHz
10MHz
100MHz
Frequency
Figure 18. μA741 frequency response, gain of 2, 1 MΩ with Cf = 5pF
Let’s move on to poles and zeros. I’ve repeated the simulation above but I increased Ci to
200 pF to separate the poles from the zero in the loop gain. Recall that a pole (zero) causes
a -90° (90°) phase shift with the pole (zero) located at -45° (45°). Looking at the phase of the
loop gain (red trace) in Figure 19, it can be seen that there are poles at 5 Hz (pole #1), 1.7
kHz (pole #2), and 1.6 MHz (pole #3) and a zero at 30 kHz (zero #1). Poles #1 and #3 are
Note that the noise gain does not continue to increase beyond the open loop gain. Once the loop gain is zero, there is no
longer any feedback. Beyond this point the noise gain is equal to the open loop gain.
20 Bob Pease is an analog guru at National Semiconductor. Bob has many good books, webcasts, and magazine articles that I
highly recommend you pursue.
19
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 17
due to the open loop gain and are seen in the yellow trace. Pole #2 and zero #1 are caused
by the feedback network21 and the locations are given by
f P2 =
1
2π (R f Ri )(C f Ci )
(28)
1
2πR f C f
(29)
f Z1 =
120
100
50
-0
Pole #1
-50
-100
Pole #2
Pole #3
Zero #1
-150
-180
1.0Hz
VDB(AvB)
VP(AvB)
10Hz
VP(Av)
VDB(Av)
100Hz
VDB(Av)-VDB(AvB)
1.0KHz
10KHz
100KHz
1.0MHz
10MHz
100MHz
Frequency
Figure 19. μA741 frequency response, gain of 2, 1 MΩ with Cf = 5pF, Cin = 200 pF
Because the noise gain is given by the reciprocal of β, it is the pole in the loop gain that
causes the peaking in the noise gain and the zero in the loop gain that reduces this peak.
Looking at Equations (28) and (29) you can see that in all cases, fZ1 ≥ fP2. As it turns out,
the maximum value of the peak is given by (1 + Ci/Cf) (32 dB in Figure 19).
Looking at Equations (28) and (29), when Rf, Ri, Cf and Ci are small, the pole and zero in
the loop gain are at (relatively) high frequencies and don’t matter. It is only when we 1)
need more bandwidth and switch to a higher frequency Op Amp such that the open loop
gain extends out to encompass fp2, 2) are forced to use high value resistors in the circuit
(e.g. when interfacing to high impedance sensors such as photodiodes), 3) have a board
layout with lots of stray capacitance, and/or 4) have to drive a large capacitive load22. The
technique applied above permits a rapid and flexible means of analyzing all of these cases
provided the design engineer is familiar with the circuit, application, and models used.
There are a few more notes to make about the frequency response plots above. The noise
gain is shown to extend out to very high frequencies. In reality this is not the case as it
Note that we are focusing on the feedback and input resistors and capacitors for this discussion. Keep in mind however that
the Op Amp output impedance, load impedance, source impedance, and strays could all have an impact on the frequency
response and stability. I’m just trying to keep the discussion simple.
22 A capacitive load combines with the Op Amp output impedance to create a pole in the loop gain which can decrease the
phase margin. Note that output impedance may not be modeled correctly so don’t assume anything.
21
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 18
does roll off with the open loop gain (although feedback factor does not for obvious reasons).
Accounting for this rolloff is important when performing a noise analysis as otherwise the
noise integral would be infinite. A more accurate representation of the noise gain can be
obtained by including the error term discussed above (see Equation (23)). This typically is
not included in a stability analysis because the angle created by the intersection of the open
loop gain curve and 1/β can be used to assess the stability of the amplifier. If the difference
between the slope of the noise gain and the slope of the open loop gain where they intersect
is less than zero, the system is stable.
For example, in Figure 14 the noise gain has a slope of zero when it intersects the open loop
gain, which has a slope of –20 dB/decade. Thus the difference is –20 dB/decade and the
system is stable (as reflected by the large phase margin). However, in Figure 15 the noise
gain has a slope nearly equal to +20 dB/decade at the intersection, so the difference is only
slightly less than 0 dB/decade. While still stable, the amplifier has considerable ringing
and a very low phase margin. Note that the locations of the poles in the open loop gain can
be obtained from the data sheet for the Op Amp and the locations of the poles and zeros in
the noise gain can be obtained by a simple circuit analysis (this is how I came up with
Equations (28) and (29)). Thus, without using PSpice, asymptotic plots of the magnitude
response can be drawn and, while not completely precise, they will reveal potential
problems without having to compute the phase response. I like that. More importantly, it is
pretty easy to remember Equations (28) and (29) such that, with a little experience, you can
almost do the analysis in your head. Note that I have neglected to include the effect of
output impedance and load capacitance. This adds another pole to the feedback factor (a
zero in the noise gain – bad news!). This can be ignored if the load capacitance is low (but
watch high frequency applications...Op Amp output impedance is not constant with
frequency.
In the discussions above, β, noise gain, bandwidth, stability, are all small signal
characteristics of an Op Amp. They matter even if you are not planning on using the Op
Amp for AC signals as the noise inherent in the Op Amp is sufficient to initiate oscillation.
However, Op Amps are frequently used to produce signals that do not meet the small signal
criteria23. The point of understanding small signal amplitude is to know the point at which
the output may start being distorted by large signal characteristics such as output drive
capability or slew rate limiting.
Output drive is straightforward; there is a limit to how much current an Op Amp output
can source (sink). Most of the specifications in the datasheet are based on the Op Amp
driving a load of a few kΩ24. Problems can arise when the output has a significant
capacitive component as the impedance of the capacitor in parallel with the resistive load
may reduce the load impedance to the point where the output drive capability of the Op
Amp is exceeded.
This is the question of how small is small? Usually the manufacturer’s data sheet will give you some
information as to the conditions under which small signal specifications are tested. An output signal amplitude
of 1V is not uncommon, although some amplifiers are specified for outputs of 100mV. You need to always read
the entire datasheet and sometimes you need to use your best judgment.
24 As an aside it is important to note that certain analog components (e.g. analog multipliers) require a fairly
specific load resistance for accurate operation. Others just need a load that is neither too small (requiring a
high current) or too large (poles introduced by strays as discussed above). For Op Amps, loads of 1 kΩ to 100 kΩ
are generally safe.
23
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 19
In the discussion of ideal characteristics at the beginning of this document it is stated that
slew rate is a large signal parameter and a measure of how fast the output voltage of an Op
Amp change. Put another way, the slew rate for the Op Amp must be greater than the
maximum slope of the output waveform over the range of frequencies being examined. If
this condition is met, the response will be determined by the bandwidth and not by slew
rate limiting.
Start to think about it this way. For small signal operation, the output is limited by the
small signal bandwidth of the Op Amp. As the output signal amplitude increases, at some
point the Op Amp will transition to where large signal parameters like slew rate begin to
dominate. Take the output sine wave
vo (t ) = V peak sin (2πft )
(30)
The maximum slew rate for this signal is given at the zero crossing, or:
dvo (t )
= 2πfV peak
dt MAX
(31)
Thus, the slew rate required for an undistorted output is a function of the output amplitude
and the frequency of the signal. Thus, the minimum25 slew rate specification for the Op
Amp must meet the following criteria to avoid slew rate limiting:
SRmin > 2πfV peak
(32)
There is a special case where we set the maximum slew rate of the output sinusoid equal to
the specified slew rate of the Op Amp. This is called the full-power bandwidth. It is a large
signal parameter and is defined at the highest frequency “full-amplitude” sinusoid that can
be output without slew rate distortion. Again, “full-amplitude” is a relative term. Usually
a 20 VPP sinusoid is used. This was because you typically could not get closer than about
two diode drops to the rail and it worked for both ±15 V and ±12 V supplies, which were
common. However, now with rails at 5V or 3.3V or less, the definition has to adapt. Again,
use your judgment and read the manufacturer’s data sheet to determine how they test this
specification. Adjusting the equation above,
f FPBW =
SR
2πV peak
(33)
If you exceed the frequency given by the equation above, the output sine wave will quickly
become a triangle wave.
So far we have been discussing the effect of these small- and large-signal parameters on
sinusoidal signals. For square wave outputs, slew rate limiting and bandwidth can both
affect the output waveform. Keep in mind that a square wave has essentially an infinite
slope. In reality, the input square wave has a slope much larger than what can be
represented at the output of the amplifier. What you will get out depends on which
specification dominates; bandwidth limiting or slew rate limiting.
25 Use caution when looking at specifications in the data sheet. “Typical” means just that. It is not a guarantee
and minimum or maximum specifications should be used to insure your design operates properly under all
conditions.
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 20
First, look at the step response due to the bandwidth. Recall that the bandwidth causes a
single pole in frequency response of the amplifier. The pole is at
ωo = 2πfT β
(34)
where 1/β is the amplifier noise gain and fT is the gain-bandwidth product. The response of
a first-order system to a step (square wave) is:
Vo (s ) =
ωoVmax
s (s + ω o )
(35)
This results in an exponential response of the form:
vo (t ) = Vmax (1 − e −ωot )
(36)
The output will look like an exponential with an initial slope of
dvo
dt
= Vmaxωo = Vmax 2πfT β
(37)
MAX
If the Op Amp slew rate is greater than this value, the exponential response will be
preserved and bandwidth limiting dominates. This is usually the case when Vmax is small
or 1/β is large.
If you instead want to measure the slew rate of the Op Amp, you would use a small gain
(typically unity) with a large voltage step (like 10 V) big enough to cause the Op Amp slew
rate specification to be much SMALLER than this slope. The result will be a linear ramp
with slope equal to the Op Amp slew rate. Keep in mind that in one case, we are still
looking at a small signal response (exponential response due to bandwidth limiting) and in
the other, a large signal parameter (slew rate limiting).
Before I wrap-up, this is a good point for an aside about making measurements on Op Amp
circuits. First, from the discussion on stability above, I hope it is clear that probing the
node connected to the inverting input is nearly pointless. Connecting a scope probe or
DMM lead to this input adds considerable capacitance and alters the noise gain, making
any signal at the output of the Op Amp meaningless. I say “nearly” because there is one
thing that can be learned by measuring the voltages at the inverting and non-inverting
inputs. If the Op Amp has the proper supply voltages and the voltage difference between
these nodes is not zero, the Op Amp is not functioning properly. Often the culprit is a bad
connection in the feedback loop, but there can be other causes as well.
While sinusoids are great test signals, they are not all inclusive. Always apply an input
step or square wave to your circuit and look at the rise time of the exponential and the slew
rate as discussed as above. This will also reveal any possible stability issues (in the form of
ringing) in your amplifier configuration.
The rise time of the exponential can provide considerable insight into your circuit provided
you are careful in setting up the measurement. Rise time is defined as the time it takes for
a signal to rise (or fall for fall time) from 10% to 90% of its final value. There is a clearly
defined relationship between the rise time and the bandwidth of an amplifier. However,
you need to use caution to be certain you are measuring the performance of your circuit and
not the limitations of your test equipment.
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
Page 21
The relationship between rise time and small-signal bandwidth is given by:
trise =
0.35
f3dB
(38)
Recognizing that for a simple RC circuit f3dB = (2πRC)-1, this is equivalent to:
trise = 2.2RC
(39)
These equations should be committed to memory, as it is extremely useful in both design
and debugging. These equations assume the signal is being affected by a single dominant
pole in the transfer function, as is typically the case in Op Amp circuits. However, in the
case where the signal passes through cascaded transfer functions (such as an amplifier
chain or an amplifier under test, scope probe, and scope amplifier), the overall rise time can
be approximated by:
2
2
t rise ≈ 1 .1 t rise _ 1 + trise _ 2 + ... + trise _ n
2
(40)
More commonly, the bandwidths of the individual transfer functions are known, or
Equation (40) can be rewritten as26:
f3dB ≈
1
1.1 f3dB _ 1
−2
+ f3dB _ 2
−2
+ ... + f3dB _ n
−2
(41)
Equation (38) can be derived as follows. Start with the transfer function for a single pole
system, H(s).
H ( s) =
k
s + ω0
(42)
Given that the input is a step function:
vi (t ) = u(t )
(43)
1
s
(44)
k
s( s + ω0 )
(45)
Vi ( s) =
Therefore:
Vo ( s) =
Applying partial fraction expansion
k
A
B
= +
s ( s + ω0 ) s s + ω0
A=
k
s + ω0
=
s =0
k
ω0
= k0
Note that some references leave out the factor of 1.1 but for most applications it improves the agreement
between the approximation and reality.
26
Copyright © 2006 Tim J. Sobering. All rights reserved.
(46)
(47)
Using Op Amps Successfully
Page 22
B=
k
s
=−
s = −τ −1
k
ω0
= − k0
k
k
k
= 0− 0
s ( s + ω0 ) s s + ω0
(48)
(49)
Taking the inverse Laplace transform yields:
v0 (t ) = k0 (1 − e − at )
(50)
tr = t0.9 − t0.1
(51)
Rise time is defined as:
Where
t0.9 = time at which v0 (t ) reaches 90% of its steady state value
t0.1 = time at which v0 (t ) reaches 10% of its steady state value
Using Equation (50)
v0 (t0.9 ) = 0.9k0 = k0 (1 − e −ω 0 t 0.9 )
(52)
0 .9 = 1 − e − ω 0 t 0 . 9
(53)
− ω 0 t 0.9 = ln 0.1
(54)
t0.9 = −
ln 0.1
(55)
ω0
v0 (t0.1 ) = 0.1k0 = k0 (1 − e−ω 0 t 0.1 )
t0.1 = −
ln 0.9
t r = t 0.9 − t 0.1 = −
tr =
1
ω0
(57)
ω0
ln 0.1 ln 0.9
+
ω0
(56)
ω0
(ln 10 − ln 109 )
(58)
(59)
Note that ω0 = 2πf 3dB . Substituting yields
tr =
ln10 − ln 109 ⎛ 1
⎜⎜
2π
⎝ f3dB
⎞
⎟⎟
⎠
(60)
or
tr =
0.35
f 3dB
Copyright © 2006 Tim J. Sobering. All rights reserved.
(61)
Using Op Amps Successfully
Page 23
The approximations in Equations (40) and (41) can be verified using the Laplace modeling
block in PSpice.
So what is the point of Equations (38), (39), (40), or (41)? One very useful application comes
in when using an oscilloscope. Let’s assume we want to look at the clock in a digital
system. Do we care if it is a 1 MHz clock, 10 MHz clock, or 100 MHz clock? In reality, the
frequency of the clock is unimportant. What matters is the rise time! Modern logic families
have rise times on the order of 1 ns. Using Equation (38) this corresponds to a signal
bandwidth of
f3dB =
0.35
= 350 MHz
1ns
(63)
When viewed on the scope display, the rise time observed will be a combination of the rise
times of the signals being measured, the oscilloscope probe, and the oscilloscope preamp, as
given by:
trise ≈ 1.1 trise ( Scope _ Amplifier ) 2 + trise ( Scope _ Probe) 2 + trise ( Signal ) 2
(62)
Assume that a 500 MHz oscilloscope and a 500 MHz probe are available for making this
measurement. The rise time observed will be closer to 1.55 ns, a ~50% error. For this
reason, it is very useful to compute the best-case rise time for the scope/probe combination
you are using and keep this in mind as you are making measurements. If the apparent rise
time of the signal you are viewing approaches the best-case value, your scope is limiting the
measurement and there may be features in the signal that are hidden from you.
The material above may seem a little overwhelming the first time you read it. No doubt
your brain glazed over a few times. This document was not intended to answer all of your
questions or address all of the conditions where Op Amps can misbehave. The intention is
to plant some seeds and illustrate the techniques for insuring that an Op Amp circuit will
function properly, or conversely, to show you where to look if a circuit isn’t working as
expected. Now you need to start playing with the simulations and, most important,
building circuits.
Addendum
As is usually the case, thoughts come to you after you think you have completed a task.
Rather than continue to revise the material above, I have decided to add notes below to
cover additional pointers for using Op Amps. Eventually this material will be incorporated
into the discussions above.
1.
Bias Currents: Another common error is to neglect to provide a DC path for the input
bias currents. For example, if you AC couple the input to a non-inverting amplifier,
you must place a resistor to ground from the non-inverting input or the bias current
will charge the coupling capacitor. The result is that after a time, the output of the
amplifier will saturate. This usually is not a problem with the inverting input as it is
either DC coupled to the output, or in the case of an integrator, DC coupled to the
input. However, I have seen designers combine integration and differentiation
functions in a single amplifier. Be sure to provide a DC path (most likely to ground).
The drawback in all cases is that adding a bias resistor can add noise to the circuit in
the form of the thermal noise of the resistor and the voltage developed by that inputs’
noise current. Everything has its price!
Copyright © 2006 Tim J. Sobering. All rights reserved.
Using Op Amps Successfully
2.
Page 24
Output Impedance – It is not always safe to assume that the output impedance of an
Op Amp is low. As the amplifier’s open loop gain decreases, it’s closed loop output
impedance increases. This can cause interesting effects when the Op Amp is
configured as an inverting amplifier because there is a direct signal path from the
input to the output (often a DC path). There is a particular issue with Sallen-Key (aka
VCVS) active filter configurations. For example, in a low pass filter using this
configuration, the response will roll off as expected and then actually rise until the
open loop gain reaches 0 dB. At this point the response flattens out. Not what one
expects from a low pass filter. When selecting an amplifier for a Sallen-Key filter, pay
close attention to the output impedance and open loop gain as you can minimize (but
not eliminate) this problem. Most designers put a passive RC filter on the output but
use caution as this can adversely affect the phase response of the amplifier. While
there are other active filter topologies, the Sallen-Key configuration is popular as it
lends itself to single supply design because it is non-inverting.
Copyright © 2006 Tim J. Sobering. All rights reserved.